3.1, 3.2, 3.3, 2.1,1.3, 3.3, 3.1
In this given data 3.1, 3.3 occurs twice
\(\therefore\) mode = 3.1 and 3.3(bimodal)

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 145+148+142+141+139+140 }{ 6 } =\frac { 855 }{ 6 } \)
x = 142.5 mm²

Total number of students = 10 + 12 + 8 + 3 = 33
The total score of 33 students = 10\(\times\)75 +12 \(\times\) 60 + 8\(\times\)40 + 3\(\times\)30
= 750 + 720 + 32090 = 1880
Mean of their score = \(\frac { Total\ Marks }{ number\ of\ students } =\frac { 188 }{ 33 } \)
= 56.96 or 57 approximately

\(\bar { x } =60kg\)  
\(\bar { x } =\cfrac { \Sigma x }{ n } =\cfrac { 56+68+72+x }{ 4 } =60\)
196 + x = 260
x = 240-196
\(\therefore \) The weight of the fourth number = 44 kg

Mean \(\bar { x } =\cfrac { \Sigma x }{ n } \)
= \(\cfrac { 26+24+28+31+30+26+24 }{ 7 } =\cfrac { 189 }{ 7 } \)
Mean temperature of the week = 27⁰C

The data given is11,15,17, x+1, 19, x-2, 3
\(\bar{x}=4\)
\(\bar{x}=\cfrac{\Sigma x}{n}=\cfrac{11+15+17+x+1+19+x-2+3}{7}\)
=\(\cfrac{66+2x-2}{7}=\cfrac{64+2x}{7}\)
\(\cfrac{64+2x}{7}=14\)
64 + 2x = 98
2x =  98 - 64 = 34
\(x=\cfrac{34}{2}=17\)
The data 11, 15, 17, x+1, 19, x-2, 3
= 11, 15,17, 18, 19, 15, 3

31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32
Writing in ascending order we get
27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 36
Number of values = 13 which is an odd number
\(\therefore\) Median =\(\left( \cfrac{13+1}{2} \right) \) Value
= \(\left(\cfrac{14}{2}\right)^{th}\) Value = 7^th value
= 32

11, 12,14,18, x+12, x+4, 30, 32, 35, 41
The number of values = 10,which is an even number
Median = Average of,\(\left( \cfrac { 10 }{ 2 } \right) ^{ th }\) and \(\left( \cfrac { 10 }{ 2 } +1 \right) ^{ th }\)
\(24=\cfrac{x+2+x+4}{2}\)
\(24=\cfrac { 2x+6 }{ 2 } =\cfrac { 2\left( x+3 \right) }{ 2 } \)
\(\therefore\) x+3 = 24
x = 24-3 = 21

47, 53, 62, 71, 83, 21, 43, 47, 41
Ascending order = 21, 41, 43, 47, 53, 62, 71, 83
The number of value 9,which is odd
\(\therefore\) Median =\(\left( \cfrac { 9+1 }{ 2 } \right) ^{ th }\quad \)
\(=\left( \cfrac { 10 }{ 2 } \right) ^{ th }\) Value = 5^th Value = 47

\(\bar { x } =20.2\) 
\( \bar{x}=\cfrac{\Sigma fx}{\Sigma f}\)
\(=\cfrac { 10\times 6+15\times 8+20p+25\times 10+30\times 6 }{ 6+8+p+10+6 } \)
\(20.2=\cfrac { 60+120+20p+250+180 }{ 30+p } \)
(30+p)20.2 = 610+20 p
606+20.2p = 610+20 p
20.2 p-20 p = 610-606 = 4
0.2 p = 4
\(\Rightarrow \) \(p=\cfrac { 4\times 10 }{ 0.2\times 10 } =\cfrac { 40 }{ 2 } =20\)

The five integers are 3, 4, 6, 9, x
Mean \(\bar{x}=\cfrac\Sigma{x}=\cfrac{3+4+6+9+x}{5}\)
=\(\cfrac{22+x}{5}\)
Their Median = 6
Mean = twice of the median
\(\cfrac{22+x}{5}=2\times6\)
22+x = 5 \(\times\) 12 = 60
\(\therefore\) x = 60- 22 = 38

Calculate the median

  N = 50
Median Class = \(\left(\cfrac{N}{2}\right)^{th}\)Value
=\(\left(\cfrac{50}{2}\right)^{th}\) Value = 25^th Value
Median value = 30 - 40
\(\cfrac{N}{2}\)=25, l = 30
m = 24, c =10, f =10
\(\therefore\) Median = \(\cfrac{l+\left(\cfrac{N}{2}-m\right)}{f}\times{c} \)
=\(\cfrac{30+25-24}{10}\times{10}=31\)

Let Assumed mean be 25

Mean \(\bar{x}=A+\frac{\Sigma fd}{\Sigma f}\)
= \(25+\frac{270}{63}\)
= 25 + 4.29 = 29.29